Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 做完了链表系列,这几天开始做树系列.对于解决关于树的问题,递归是一个很…

题目链接:Balanced Binary Tree | LeetCode OJ Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more tha…

Problem Link: http://oj.leetcode.com/problems/balanced-binary-tree/ We use a recursive auxilar function to determine whether a sub-tree is balanced, if the tree is balanced, it also return the depth of the sub-tree. A tree T is balanced if the follow…

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题目: Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 提示: 此题要求判断一个给定的二叉树是否是“平衡”的,这里“平…

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 就是去查看一棵树是不是平衡的,一开始对平衡二叉树的理解有错误,所以写错…

剑指offer 面试题39:判断平衡二叉树 提交网址:  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId=13&tqId=11192 时间限制:1秒       空间限制:32768K      参与人数:2481 题目描述 输入一棵二叉树,判断该二叉树是否是平衡二叉树. 分析: 平衡二叉树定义 递归解法 AC代码: #include<iostream> #include<vector&…

110.Balanced Binary Tree Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 很早以前做的了  准…

Question 110. Balanced Binary Tree Solution 题目大意:判断一个二叉树是不是平衡二叉树 思路:定义个boolean来记录每个子节点是否平衡 Java实现: public boolean isBalanced(TreeNode root) { boolean[] balanced = {true}; height(root, balanced); return balanced[0]; } private int height(TreeNode node,…

Balanced Binary Tree [数据结构和算法]全面剖析树的各类遍历方法 描述 解析 递归分别判断每个节点的左右子树 该题是Easy的原因是该题可以很容易的想到时间复杂度为O(n^2)的方法.即按照定义,判断根节点左右子树的高度是不是相差1,递归判断左右子树是不是平衡的. 根据深度判断左右子树是否平衡 在计算树的高度的同时判断该树是不是平衡的. 即,先判断子树是不是平衡的,若是,则返回子树的高度:若不是,则返回一个非法的数字,如负数. 当一个节点是左右子树有一个不是平衡二叉树则不必继…